Existence of an identity element,

For example, let Σ = {a, b, c} and Σ* = {ε, a, b, c, a b, b c, c a, …….} is the set of all possible strings form over the alphabets Σ then algebraic system (Σ*, .) is monoid, where . is a binary concatenation operation, i.e., for any two strings x, y ∈ Σ*, x.y yields the string xy.

Since,

1. For any x any y ∈ Σ*, x . y ⇒ xy ∈ Σ*; hence operation . is closed.

2. For any x, y, and z ∈ Σ*, (x . y) . z = x . (y . z) ⇒ x y z ∈ Σ*; so operation . is associative.

3. Set Σ* contains an identity element ε called null string i.e. | ε | = 0, then for any x ∈ Σ*,

x . ε = ε . x = x

Therefore, algebraic system (Σ*, .) is a monoid. If we replace the set Σ* by Σ⁺, where Σ⁺

= Σ* – {ε}, where set Σ⁺ contains all possible strings formed over the alphabet Σ except null string (ε) then algebraic system (Σ⁺, .) is not a monoid due to non existence of an identity element with respect to operation concatenation but a semigroup.

Operation Table

The properties hold by an algebraic system can be easily observed from the operation table.

Let (X, ) is an algebraic system, where is a binary operation on X, then operation table presented an arrangement of values resulted after performing the binary operation over the elements of set X. For example, the algebraic structure (X, ⊕) is a monoid, where ⊕ is a binary operation defined over set X = {0, 1, 2, 3}, i.e.

a ⊕ b = a + b, if a + b ≤ 3, otherwise a + b – 4 Construct the transition table for algebraic system (X, )

⊕ 0 1 2 3

0 0 1 2 3

1 1 2 3 0

2 2 3 0 1

3 3 0 1 2

Fig. 4.2 Operation table.

From the operation table shown in Fig. 4.2 we derive following conclusions,

l Since each element in table belongs to set X, hence operation ⊕ is closed.

l Operation is associative.

l Values of the first column are similar to the corresponding element of the set when operated on ⊕ over element 0, hence 0 is a unique identity element.

Therefore, algebraic system (X, ⊕) is a monoid. Further, we see that algebraic system (X, ⊕) is also a group.

l Since there is an occurrence of the identity element 0 in each row in the operation table which shows the existence of the inverse element for every element of X, i.e.,

0 ⊕ 0 = 0; 1 ⊕ 3 = 0; 2 ⊕ 2 = 0; 3 ⊕ 1 = 0.

l Also corresponding rows and column are same in the table; hence operation ⊕ holds commutative property.

Therefore, algebraic system (X, ⊕) is an Abelian group.

Example 4.2. Let X = {p, q, r} and binary operation ⊗ defines in the operation table shown in Fig. 4.3 then algebraic system (X, ⊗) is a semi group but not monoid.

Sol.

⊕ p q r

p p p p

q q q q

r r r r

Fig. 4.3 Operation table.

l Operation ⊗ is closed, since all elements in the table belong to set X.

l Operation is associative.

Hence, (X, ⊗) is a semigroup. Further,

l Algebraic structure (X, ⊗) not posses a unique identity element for all elements of X.

Here elements identity elements are {p, q, r}.

l Since corresponding rows and columns are not same hence operation ⊗ is not commu-tative.

Therefore, (X, ⊗) is not monoid and also not a group.

Example 4.3. Let X = {1, – 1, i, – i} and the binary operation H is define in the operation table (Fig. 4.4). Prove that algebraic system (X, H) is a group.

Sol.

H 1 – 1 i – i

1 1 – 1 i – i

– 1 – 1 1 – i i

i i – i – 1 1

– i – i i 1 – 1

Fig. 4.4 Operation table.

(Ready self verify that operation table holds all restrictions required by a group) Modulo Operation

Now we define two special operations called additional modulo and multiplication modulo over the set X, where

l Additional Modulo n if a, b ∈ X then a addition b modulo n is defined as, (a +_n b) = (a + b) mod n = r (where r > 0)

For example, let a = 11 and b = 6 then its addition modulo 5 is given by (11 +₅ 6) = (11 + 6) mod 5 = 2 (> 0). Consider another set of values such that a = – 15 and b = 5, then

(– 11 +₅ 5) = (– 11 + 5) mod 5 = (– 6) mod 5 = [(– 5 ∗ 2) + 4] mod 5 = 4 (> 0).

Let a = – 10 and b = – 7 then

(– 10 +₅ 7) = (– 10 + (– 7)) mod 5 = (– 17) mod 5 = [(– 5 ∗ 4) + 3] mod 5 = 3 (> 0).

l Multiplication Modulo n if a, b ∈ X then a multiplication b modulo n is defined as, (a *_n b) = (a ∗ b) mod n = r (where r > 0)

Example 4.4. Let (X, ×) be a group, where × is the multiplication operation on X, then for any x, y ∈ X prove that,

1. (x^–1)^–1 = x, and

2. (x × y)^–1= y^–1 × x^–1 or (x × y) = (y^–1 × x^–1)^–1; Sol.

1. Recall the definition of the group, such that for every element x ∈ X there exists an unique inverse say x′ ∈ X i.e.,

x × x′ = ê (identity element) = x′ × x;

From the symmetricity,

x′ = x^–1 or x′^–1 = x Replace value of x′ by x^–1 in so, we get

(x^–1)^–1 = x (Proved)

2. From the result of equality (1), for any x ∈ X, we have the inverse element x^–1, where x × x^–1

= ê; and also y × y^–1 = ê for any y ∈ X. Since, operation × is closed so x × y ∈ X, Let u = x × y and w = y^–1 × x^–1, then

u × w = (x × y) × (y^–1 × x^–1)

= x × (y × y^–1) × x^–1 (by Associativity)

= x × ê × x^–1 (since y × y^–1 = ê)

= x × x^–1

= ê (identity element) Thus we have,

(x × y) × (y^–1 × x^–1) = ê = (y^–1 × x^–1) × (x × y) also.

⇒ (x × y) = (y^–1 × x^–1)^–1, and due to symmetricity

(x × y)^–1 = (y^–1 × x^–1). (Proved)

Example 4.5. Let W = {1, ω, ω²}, where ω is cube root of unity i.e. ω³ = 1 and 1 + ω + ω² = 0, then show that algebraic system (W, ×) is a group.

Sol. Since,

1. Operation × is closed for W, i.e.

1 × ω = ω ∈ W; ω × ω² = ω³ = 1 ∈ W; 1 × ω² = ω² ∈ W; and others.

2. Operation × is associative over W, viz.

ω² × (1 × ω) = (ω² × 1) × ω = ω³; and others.

3. Existence of an identity element i.e., ω × ê = x = x × ê for any x ∈ W

if x = ω then ê = 1, i.e. 1 × ω = ω × 1 = ω;

if x = ω² then ê = 1, i.e. 1 × ω² = ω² × 1 = ω²;

4. Existence of unique inverse element for every element of W, i.e., x × y = y × x = ê where x, y ∈ W and y is inverse of x.

if x = 1 then y = 1, so 1 × 1 = ê;

if x = ω then y = ω², so ω × ω² = ω³ = 1 = ê;

if x = ω² then y = ω, so ω² × ω = ω³ = 1 = ê;

Therefore, Algebraic system (W, ×) is a group.

4.3 SEMI SUBGROUP

Let algebraic system (X, ) is a semi group and let Y ⊆ X, then (Y, ) is said to be a semi subgroup if, (Y, ⊗) is itself a semigroup, i.e.

(Y, ⊗) ⊆ (X, ⊗)

For example, (I⁺, +) is semigroup, where binary operation + is defines over set of posi-tive integers I⁺. Now consider two subsets of I⁺ i.e., (1) positive integers that are all even (J) and (2) positive integers that are all odd (K). Thus,

J ⊆ I⁺ and K ⊆ I⁺

Now test whether (J, +) is a semigroup or (K, +) is a semigroup. Former, is a semigroup because operation + is closed and associative over J but latter is not a semigroup due to violation

of closure property by operation + over K (since addition of two positive odd integers must be group (X, ×). Out of these complexes some complexes may form a group or may not form a group under bounding operation ‘×’.

4.5 PRODUCT SEMIGROUPS

Let (X, ⊕), and (Y, ⊗) are two semigroups, define Z = X × Y such that the elements of Z are ordered pair (x, y) where x ∈ X and y ∈ Y. Then (Z, #) is also a semigroup by assuming (a, b)

∈ Z and (c, d) ∈ Z i.e., (a, b) # (c, d) = (a ⊕ c, b ⊗ d). Hence, (Z, #) is called a product semigroup.

Consider an example, algebraic system (R⁺, ×) where × is usual multiplication operation defines over the set of positive real numbers R⁺ forms a semigroup. Similarly (I, +₄) where +₄ is addition modulo 4 operation define over set of Integers forms a semigroup. Let S is the direct product, i.e., S = R⁺ × I, then check whether (S, #) is a semigroup or not. Assume ordered pairs (a, b) ∈ S and (c, d) ∈ S, where elements a, c belongs to R⁺, and elements b, d belongs to I then

(a, b) # (c, d) = (a × c, b +₄ d)

Since operation × is closed in R⁺ and operation +₄ is closed in I, hence operation # is closed. Also, since operation × and +₄ are associative hence # also holds associativity. Therefore, direct product is a semigroup.

4.6 PERMUTATION GROUPS

Let X be a group and p : X → X is a mapping to be permutation of X if p is bijective (one-one and onto). Such groups are called permutation groups. Let X = {a, b, c,…..} be a set and let p denotes a permutation of the elements of X; i.e., p : X → X is a bijective, then

p =

F

_{p a( )}^a _{p b}^b_{( ) p c( )...}^c...

HG I

KJ

is called permutation group, where image of the element of X is entered below the element.

For example, let X = {1, 2, 3} and suppose p(1) = 2, p(2) = 3, p(3) = 1 then we may represent p as,

and so on. In general, a group of n-elements can have a total of n ! permutations that is called order of permutation and the degree of permutation is the number of the elements of the set X. In the said example, the degree of permutation is 3 and the order of permutation is 6.

Let p₁, p₂, …….p_n! are all possible permutations of a set of n elements then it is called symmetric set of permutations. In the previous example {p₁, p₂, …p₆} are the symmetric set of permutations that are shown below.

p₁ = 1 2 3

Let p be the permutation where, p = a a a b₁¹ b²₂ b₃³

F HG I

KJ

then, inverse of permutation is denoted by p^–1 and is defined as, p^–1 = ^b_a¹ _a^{b2 b}_a³

1 2 3

F HG I

KJ

Similarly, a permutation is said to be an identity permutation if image of every element of set X is same to the corresponding element, i.e., if X = {a, b, c} then p(a) = a, p(b) = b, and p(c) = c, then permutation p where,

p =

F

^{a b c}_{a b c}

H I

K

is called an identity permutation.

4.7 ORDER OF A GROUP

Let (X, ) be a group, then order of the X is denoted by O(X) is the number of elements in group X. For example if group X consisting of m elements then O(X) = m. If X is infinite then O(X) = ∞.

Let x ∈ X then order of an element x, denoted as O(x) is n if aⁿ = e (identity element).

For example, consider X = {1, – 1, i, – i} is a group under usual multiplication operation × i.e.

(X, ×).

Then order of the group is, i.e., O(X) = 4. The order of its elements is determined as follows : order of the group X and order of its elements.

Sol. Since, set X contains three elements hence, O(X) = 3. The order of the elements is deter-mined as follows :

From the operation table shown in Fig. 4.5 the group (X, ×) has an identity element 1 Therefore,

l O(1) = 1 [∵ 11 = 1 (identity element)]

l O(ω) = 3 [∵ (ω)³ = 1 (identity element)]

l O(ω²) = 3 [∵ (ω²)³ = 1 (identity element)]

× 1 ω ω²

1 1 ω ω²

ω ω ω² 1

ω² ω² 1 ω

Fig 4.5 Operation Table.

4.8 SUBGROUPS

Let algebraic system (X, ) is a group, where X be a nonempty set and be a binary operation on X. A subset Y of X such that (Y, ) is said to be a subgroup of X if (Y, ) is also a group together following conditions are satisfie,

1. is closed, 2. is associative,

3. Existence of an identity element ê ∈ Y, where ê is the identity element of (X, ), and 4. Existence of unique an inverse for (X, ) and also for (Y, ).

We can use the following notation for a subgroup,

Y < X, where (Y, ) is the subgroup of (X. ).

For example, algebraic system (I, +), where I is the set of integers and operation + is usual addition operations of integers, is a group. Also, I⁺ ⊆ I (where I⁺ is the set of all positive integers) and since, (I⁺, +) is a group and satisfy all the restrictions 1 – 4 therefore, (I⁺, +) is a subgroup.

We further define, if Y and Z are two subsets of a group X, then

l YZ = (yz/y ∈ Y, z ∈ Z} and

l Y^–1 = {y^–1/y ∈ Y}; inverse of Y is also the subset of group X.

l We define, Y^k (for k = 0, 1, 2, ….) i.e.

Y¹ = Y; Y² = YY; ……..; similarly Y^k+1 = Y^k Y; and Y⁰ = {ê}

where,

Y² = {y₁y₂/y₁, y₂ ∈ Y}, and so on.

The subset Y^–k is defined as (Y^–1)^k. In particular, we write, for x ∈ X,

Y x = {yx/y ∈ Y} or, x Y = {xy/y ∈ Y}

If Y ≠ Ø, then we can see that

Y X = X Y = X; X^–1 = X; X ê = ê X = ê;

Example 4.7. Let Y is a subgroup of X, i.e. Y < X if and only if YY^–1 ⊂ Y.

Sol. YY^–1 ⊂ Y ⇒ if a, b ∈ Y then ab^–1 ∈ Y.

Necessity is obvious.

For Sufficiency,

if a ∈ Y ⇒ aa^–1 ∈ YY^–1 ⊂ Y;

⇒ aa^–1 ∈ Y, i.e. ê ∈ Y.

So, ê a^–1 ∈ Y Y^–1 ⊂ Y, thus a^–1 ∈ Y.

Similarly, if a, b ∈ Y ⇒ b^–1 ∈ Y and ab^–1 ∈ Y;

So, (ab^–1)^–1 ∈ YY^–1 ⊂ Y, i.e. a, b ∈ Y

Thus, a ∈ Y ⇒ a^–1 ∈ Y and a, b ∈ Y ⇒ a b ∈ Y, hence Y becomes a subgroup.

Example 4.8. Let Y and Z are two subgroups of X, then product of YZ is a subgroup of X if and only of YZ = ZY.

Sol. Necessary Condition. Since, Y, Z, and YZ are subgroups of X, i.e.

Y < X, Z < X, and YZ < X Therefore, Y^–1 = Y, Z^–1 = Z, (YZ)^–1 = YZ, But (YZ)^–1 = Z^–1Y^–1, = ZY

Thus, ZY = YZ

(Sufficient condition) To claim YZ < X, from given Y < X, Z < X, and YZ = ZY Since, we have Y² = Y = Y^–1 and Z² = Z = Z^–1

Assume S = YZ

Then, S² = (YZ) (YZ) = Y (ZY) Z = Y (YZ) Z (Commutivity)

= (YY) (ZZ)

= Y² Z²

= YZ (∴ Y² Z² = YZ)

= S

also, S^–1 = (YZ)^–1 = Z^–1 Y^–1

= ZY

= YZ (∴ ZY = YZ)

= S

So, we have S² = S = S^–1, therefore S is a subgroup of X, or YZ < X.

4.9 CYCLIC GROUPS

A group (X, x) is said to be cyclic group if ∀x ∈ X, there exists a, fixed element g such that, gⁿ= x for some integer n, where g is called generator of the cycle. Alternatively, a group X is said to be cyclic with generator g if every element of X is in g*, where g* is of the form,

g⁰ = e; g¹ = g; g² = g × g; ………….; gⁿ = g × g × g ……… × g (n times) (where n ∈ I)

l Consider an example of group (X, ×), where X = {1, – 1, i, – i}. Since every element of X is generated from one of the element ‘i’ of the set, i.e.,

i⁴ = 1; i² = – 1; i³ = i; i⁴ = – i.

Since each element can be expressed in some power of i so i is one of the generator of the cyclic group, hence,

g = [i]

Similarly, other element ‘– i’ also generates all the elements of the set X, hence other generator, i.e., g = [– i], where -i is the inverse element of i.

Hence, we can say that if a is the generator of the cyclic group X then a^–1 is also a generator of X.

Therefore, g = [i] or [– i] are two generators of the cyclic group X.

l Consider another example of group (X, ×), i.e. X = {1, ω, ω²}, where ω is cube root of unity. Then group is cyclic with the generator g = [ω] because, the elements of X are expressed in ω³, ω, ω² respectively for 1, ω, and ω². Further, the inverse element of ω i.e., ω² is also the generator. Therefore, g = [ω] and [ω²].

Example 4.9. Show that group (I, +) is a cyclic group. Find its generator (where I is the set of integers)

Sol. Since, the set I = {……– 2, – 1, 0, 1, 2, ……}. The elements of I can be generated from on of the element ‘1’ i.e., n = 1 + 1 + ……+ 1 (n times) = n. g. Since + is closed operation on generator g, so we can write as, g + g + ……. + g (n times) = n. g, where,

(1)⁰ = 0 (identity element), (1)¹ = 1, (1 itself)

(1)² = 1 + 1 = 2,

(1)³ = 1 + 1 + 1 = 3, and so on.

Similarly,

(1)^–1 = [(1)¹]^–1 = [1]^–1 = – 1,

(1)^–2 = [(1)²]^–1 = [2]^–1 = – 2, and so on.

Therefore, g = [1] and also [– 1].

Example 4.10. Show that group X = ({1, 2, 3, 4, 5, 6}, ×₇) is a cyclic group, where ×₇ is a multiplication modulo 7 operator.

Sol. Since the elements of the group X are generated from the element 3 and 5, and there generations are shown below.

(3)⁰ = 1 (identity element), Also, (5)⁰ = 1 (identity element), (3)¹ = 3, (generator itself) (5)¹ = 5, (generator itself) (3)² = 3 ×₇3 = 2, (5)² = 5 ×₇5 = 4,

(3)³ = 3 ×₇ 3 ×₇ 3 = 6, (5)³ = 5 ×₇ 5 ×₇ 5 = 6, (3)⁴ = 3 ×₇ 3 ×₇ 3 ×₇ 3 = 4, (5)⁴ = 5 ×₇ 5 ×₇ 5 ×₇ 5 = 2, (3)⁵ = 3 ×₇ 3 ×₇ 3 ×₇ 3 ×₇ 3 = 5 (5)⁵ = 5 ×₇ 5 ×₇ 5 ×₇ 5 ×₇ 5 = 3 Therefore, g = [3] and [5].

(here 5 is the inverse element of 3) Facts

l We can denote the cyclic group X generated by g as, X = <g>

l A cyclic group X = <g> of order m can be define as, i.e., it consists of the powers, X = <g> = {g⁰, g, g², ……., g^m–1}, where g^m = ê;

l Every subgroup of a cyclic group is cyclic.

l If a is the generator of the cyclic group X then inverse element of a is also the gen-erator of X.

4.10 COSETS

For a group (X, ) let (Y, ) is a subgroup of X (i.e. Y ⊆ X), then we define the cosets of Y for any arbitrary x ∈ X are,

(Left coset) x Y = {x y/y ∈ Y}, and (Right coset) Y x = {y x/y ∈ Y}

Left cosets and right cosets may or may not be equal. They are equal only for commuta-tive groups, otherwise they are unequal.

For example, let I be the additive group of integers, i.e. (I, +) now take a subset of I is J where J = 3I, so J = {……, – 6, – 3, 0, 3, 6, ….} [J ⊆ I]

Thus cosets of J in I generated by element 0, 1, 2 (∴ 0, 1, 2 ∈ I) are correspondingly given as,

0 + J = { ……, – 6, – 3, 0, 3, 6, ….}, 1 + J = {……, – 5, – 2, 1, 4, 7, ..….}, 2 + J = {……, – 4, – 1, 2, 5, 8, ..….}, 3 + J = {……, – 6, – 3, 0, 3, 6, ….} = 0 + J 4 + J = {……, – 5, – 2, 1, 4, 7, ..….} = 1 + J 5 + J = {……, – 4, – 1, 2, 5, 8, .…..} = 2 + J 6 + J = {……, – 6, – 3, 0, 3, 6, .…..} = 0 + J and so on, hence cosets,

0 + J = 3 + J = 6 + J = …… = { ……, – 6, – 3, 0, 3, 6, ….}

1 + J = 4 + J = 7 + J = ….. = {……, – 5, – 2, 1, 4, 7, ..…}

2 + J = 5 + J = 8 + J = …… = {……, – 4, – 1, 2, 5, 8, .…}

From these cosets we can easily find the entire set I, i.e., I = (0 + J) ∪ (1 + J) ∪ (2 + J)

Therefore, cosets are the way of partitioning the entire set into smaller sets.

0 + J 1 + J

2 + J I Fact

Let X is a group and Y is its subgroup then, number of distinct cosets of Y is called the index

In document Fundamentals of Mathematics for Computer Science (Page 97-106)